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if [tex]g(s)= \sum_{n=1}^{\infty} a(n) n^{-s}[/tex]
Where g(s) has a single pole at s=1 with residue C, then my question/conjecture is if for s >0 (real part of s bigger than 0) we can write
[tex]g(s)= C(\frac{1}{s-1}+1)-s\int_{0}^{\infty}dx(Cx-A(x))x^{-s-1}[/tex]
of course [tex]A(x)=\sum_{n \le x}a(n)[/tex]
the question is if the series converge for s >1 with a pole there is a method to 'substract' this singularity (pole) at s=1 to give meaning for the series at any positive s.
I think that the 'Ramanujan resummation' may help to give the result:
[tex]\sum_{ n >1}^{[R]}a(n)n^{-s} = g(s)-C(s-1)^{-1}[/tex] valid even for s=1 or s>0 (??)
Where g(s) has a single pole at s=1 with residue C, then my question/conjecture is if for s >0 (real part of s bigger than 0) we can write
[tex]g(s)= C(\frac{1}{s-1}+1)-s\int_{0}^{\infty}dx(Cx-A(x))x^{-s-1}[/tex]
of course [tex]A(x)=\sum_{n \le x}a(n)[/tex]
the question is if the series converge for s >1 with a pole there is a method to 'substract' this singularity (pole) at s=1 to give meaning for the series at any positive s.
I think that the 'Ramanujan resummation' may help to give the result:
[tex]\sum_{ n >1}^{[R]}a(n)n^{-s} = g(s)-C(s-1)^{-1}[/tex] valid even for s=1 or s>0 (??)
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